Bolted Connections

Bolted Connections- Non-moment

Bolted connections are very popular in steel structures. In this article I will discuss about the bolted connection designs. Answers will be provided to the questions like, How does a bolted connection work? Structural bolts are utilised basically in Shearing, Tension and in their combinations. The combination of shearing and tension is the mostly used case in a bolted connection. I will provide explanation in simple language so that the people with limited knowledge can also get benefit from this article. Engineering students, Engineers who want to refresh their knowledge and those who want to learn structural engineering will get benefit from this article.

It is assumed that the reader has adequate knowledge of the contents that is provided in the previous article.

When we design a bolted connection, in addition to bolts, status of the connecting and connected members or plates need to be checked as well.

Checking of bolts

Bolts have two areas of interest, i.e., Shank and Core. Non threaded portion is called shank and the threaded portion is called a core. Net cross section area of the bolt at core is less than that of the shank. Therefore, shear strength of the bolt at core area is comparatively less. Since cross section area at core is reduced, the tensile strength of the bolt is computed based on the core.

Before determining strength of the bolts, the following properties of the bolts need to be collected.

-Ultimate tensile strength of the bolt, fuf. Here the last f stands for fastener.

-Net cross section area of the bolt, an. The Net cross section area used for shearing depends on the fact that if threads are included in the shear plane or not, i.e., anc or ans. Where c stands for core and s stands for shank.

Bolted connection of a 3-D roof truss system

Tensile strength of bolts depends on core area.

Tension capacity of a bolt, фNtf =0.8 x fuf x anc >= Nt*

Here, ф=0.8 is strength reduction factor, Nt* is factored tension force in a bolt

When a connection is in tension, the connecting/base plates need checking for prying caused actions as well.In order to account for the prying, the force on bolts may be increased roughly by about 20%.

Bolts in tension with possibility of prying

Following example shows how bolts in tension are checked:

Say, Factored total tension, N*= 160 kN, Bolt diameter, df= 12 mm, minimum tensile strength of bolt, fuf = 830 N/mm2

Cross section area of a bolt for tension, anc= say 0.72 x 3.14/4 x 12^2= 81.4 mm2

Tension capacity of a bolt, фNtf =0.8 x 830 x 81.4= 54049 N= 54.05 kN

Factored tension force on a bolt= 160/4= 40 kN

Add 20% for prying, Nt*= 20/100*40 + 40= 48 kN

Since, фNtf = 54 kN > Nt*= 48 kN, the connection may be accepted. However the baseplates need checking for the actions due to prying.

Shear strength of  bolts

фVf =0.8 x 0.62 x fuf x k x an >=Vf*

Depending on the case, threads included or excluded in the shear plane, an can be anc or ans. The anc is the cross section area of a bolt calculated based on core area. Whereas ans is based on shank.

Here, ф=0.8 is strength reduction factor, Vf*: factored shear force in a bolt

k: reduction factor for lapped connection. For non-lapped connection k = 1 may be used. For other cases value of "k" shall not be greater than 1.0. Larger the "L" lesser is the value of "k". Value of "k" shall be used as recommended by national code of practice.

 

Bolt in shear and Tension When bolts are in shear and tension then the bolt shall pass the following unity check:

[Nt*/фNtf ]^2 +[V*/фVf ]^2 =< 1

Checking of plates

Similar to bolts, strength of connecting and connected members/plates need checking also for bearing, tearout and tension.

Bearing Strength of plates In addition to ultimate tensile strength of the plate material, fup, the bearing strength of the plates depends on its thickness and bolt diameter.

Bearing strength of a plate, фVbb=0.9 x 3.2 x d x t x fup  >= V*

Here, ф=0.9 is strength reduction factor, V*: factored bearing force on a plate, d: bolt diameter,

t: thickness of the plate

Tearout Strength of Plates Plates in tension may fail in tearout as well. In this case, in addition to fup, the plate tearout strength also depends on the clear end distance of the plate from end of the last bolt hole. Tearing can also extend into the space between bolt holes.

фVbt=0.9 x e x t x fup  >= V*

Here, ф=0.9 is strength reduction factor, V*: factored bearing force on a plate, t: thickness of the plate

e: half diameter of the bolt plus minimum end distance of a plate from edge of the bolt hole. This distance shall be measured along the direction of force application. 

Tensile Strength of plate Tensile strength of plates are generally checked using their net cross section area.

фNtp =0.85 x fup x anp >= Ntp*

         

However the tension capacity of the plate is recommended to be checked also using the yield strength which is as follows,

фNtp =1 x fy x ag >=Ntp*

Here, фNtp: Tension capacity of a plate, 0.85 and 1: strength reduction factors, anp: net cross section area of a plate, ag: gross cross section area of a plate, fup: minimum tensile strength of plate material, fy: Yield strength of plate material

Example bolted connection of an angle section

In most of the cases when more than one rows of bolts are used, the bolts are arranged in staggered position. This arrangement helps in avoiding reduction of net cross section area of the plate. In this case the net cross section area of the plate is calculated in the following way:

an= t x [b -(3 x dh) + (Sp1^2/(4 x Sg1)+Sp2^2/(4 x Sg2))]

Consideration of staggered bolt holes

If bolt holes are in a straight line, unstaggered, the last two additive parts of the formula is not required. 

Where, t: plate thickness, b: width of plate, dh: diameter of bolt hole

Following example shows a general approach on how bolts and plates are checked in a bolted connection.

Fig.1: Example of a simple bolted connection

Let's assume following data for the connection shown in Fig.1 above:

Factored force, V*= 220 kN, Thickness of top and bottom plates, t = 6 mm, Thickness of middle plate, tm =10 mm, Diameter of bolts, d =20 mm, minimum tensile strength of bolt, fuf = 830 N/mm2, minimum tensile strength of plates, fup = 410 N/mm2, similarly, yield strength of the plates, fy= 250 N/mm2 and width of the plates, b= 90 mm, bolt hole diameter= 22 mm

We have two bolts, therefore maximum shear force in a bolt, Vf*= 220/2= 110kN 

Since, threaded portion of the bolt is excluded from the shear plane, the bolt cross section area,
an= anc= 3.14/4*20^2= 314 mm2. This case is called a "threads excluded" case. Here 3.14 is value of ㄫ. If the threaded portion was included in the shear plane, then the reduced diameter of the bolt should have been used to calculate the cross section area of the bolt. In that scenario the case is called  a "threads included" case.

Shear strength of a bolt, фVf =0.8 x 0.62 x fuf x k x an= 0.8*0.62*830*1*314= 129267 N= 129.27 kN

Here, the bolt is sheared twice (on top and bottom surfaces of the middle plate), therefore, the shear strength of a bolt is 2*129.27= 258.5 kN > 110 kN okay. Such case of shearing is called a "double shearing" case.

Since these bolts are not utilised in direct tension, we don't need to check the bolts for tension. I will show how bolts in tension are checked when they are utilised in combination with shearing.

Maximum bearing strength of a plate, фVbb=0.9 x 3.2 x d x t x fup= 0.9*3.2*20*10*410= 236160 N

= 236.16 kN (middle plate)

Similarly, bearing capacity of a plate,  фVbb= 0.9*3.2*20*6*410= 141696 N. Since the top and bottom plates are bearing a bolt, total bearing strength of two plates,фVbb= 2*141696= 283392 N= 283.39 kN

236.16 kN > 110 kN okay. Since the total tension is borne by two bearing planes, the relation may also be expressed as, 2 x 236.16=  472.32 kN> 220 kN

Tensile strength of the plates need to be checked in addition to bearing and tearout.

фNtp= 0.85 x fup x anp= 0.85*410*(90-22)*10= 236980 N= 236.98 kN (middle plate)

фNtp =1 x fy x ag= 1*250*90*10= 225000 N= 225 kN (middle plate)

фNtp= 0.85 x fup x anp= 0.85*410*(90-22)*6= 142188 N= 142.19 kN (top/bottom plate), 284.38 kN for two plates 

фNtp =1 x fy x ag= 1*250*90*6= 135000 N= 135 kN (top/bottom plate), 270 kN for two plates

Total capacity of the connection is therefore 225 kN > 220 kN okay. Therefore, the design may be accepted.

Block Shear In addition to what have been mentioned above, the plates in bolted connection must be checked also for the possible block shear failure. Strength of plates against block shear depends on bolt diameter, plate thickness, plate end/edge distances and clear distance between bolt holes in rows. Connection can fail in different modes of block shear. Checking of a connection for possible modes of block shear failure shall be carried out.

Some examples of block shear failure modes are shown below.

Some examples of block shear failure modes

In the next article, I will discuss on how moment resisting bolted connections are checked.
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